One way of formulating this question rigorously would perhaps be to ask whether the forgetful functor from the category of division rings to the category of sets had an adjoint.
–
Kevin BuzzardApr 25 '10 at 9:04

2 Answers
2

If you had a "free division ring" $F$ on a set $X$, and any division ring $R$, then any set theoretic map $X\to R$ would correspond to a unique division ring homomorphism $F\to R$. If $X$ has at least two elements $x\neq y$, let $R=\mathbb{Q}$. Then you cannot extend both a set theoretic map $f\colon X\to R$ that sends $x$ to $0$ and $y$ to $1$, and a set-theoretic map $g\colon X\to R$ that sends $x$ to $1$ and $y$ to $0$: division rings are simple, so any homomorphism must be either one-to-one or the zero map. (I put both maps, in case one wonders whether you can have $x$ correspond to the zero element of $F$). So, no, you cannot have "free division rings", much like you cannot have "free fields".

Perhaps it is more interesting to require that the set-theoretic map have image in $R^\times$, to avoid the obstruction you mention? In that case, perhaps a quotient of $\mathbb Q[X, X^{-1}]$ by a maximal ideal would do the job.
–
L SpiceApr 25 '10 at 7:37

Sorry, never mind, that was silly. In your notation, in any putative such free division ring, taking an 'extension' of the map sending $x$ and $y$ both to $1$ would show that $x - y = 0$, preventing extension of a map sending $x$ to $0$ and $y$ to $1$. Is there any possibility of an interesting replacement for the natural universal condition that you mention, or is it clear that no sensible such condition exists?
–
L SpiceApr 25 '10 at 7:45

Thanks for this answer; you certainly dispelled my high dreams. How about an answer to the now meaningless second question? Is it possible to have a finitely generated division ring with infinitely many "independent" elements? I'm not sure what independent means here, or how to formulate it exactly. The perfect characterization would be, of course, that they generate a free division ring, but that's now meaningless.
–
Uri AndrewsApr 25 '10 at 10:08

@Uri: Here's one possible way of asking that question: every division ring is a division algebra over its center (which is nontrivial, since it contains $0$ and $1$). So we can ask whether there exists a division ring $R$, and a subring $S$ that contains $C(R)$, also a division ring, such that $R$ is finitely generated as a division algebra over $C(R)$ but $S$ is not finitely generated as a division algebra over $C(R)$. Or same question but relative to $C(S)$, or some common subring that is a division algebra. I'll think about it a bit.
–
Arturo MagidinApr 25 '10 at 19:02

@Spice: I don't think you can make it sensible in the sense of getting an adjoint to the forgetful functor to sets, which is the usual meaning of "free object". I was tempted to try with "one-to-one maps into $R^{\times}$, but you will run into trouble if there are any "algebraic" relations between them; for example, mapping $x$ to $y^2$, leading to a non-trivial map that has a nontrivial kernel ($x-y^2$ would map to zero). So I think the answer is that there is no reasonable way to restrict the universal property to get a nice notion here.
–
Arturo MagidinApr 25 '10 at 19:05

There is a notion of free division ring, due to Paul Cohn. I don't know a good online reference, but this AMS bulletin article talks about the construction and gives references. They go by the name of "free skew field" rather than "free division ring".

In the usual category of division rings and ring homomorphisms, there is no free division ring in the sense of category theory, but Cohn's free division rings are pretty close. If you consider the category of division rings with specializations as morphisms, then Cohn's construction gives you precisely the free objects.

Let $D, D'$ be division rings. Then a specialization is a homomorphism $R \to D'$, where $R$ is a subring which generates $D$ as a division ring. (In other words, the smallest division ring within $D$ containing $R$ is $D$ itself.) Specializations are a reasonably natural idea. For example, let $C$ be the complex numbers, and $C(x)$ be the field of rational functions over $C$. Then all specializations between $C(x)$ and $C$ are given by sending $p(x)$ in $C(x)$ to the value $p(a)$ for some $a$ in $C$. This is not defined for all rational functions, but only for the rational functions that don't have a pole at $a$.

If you restrict to the category of fields over a fixed base field $k$ and specializations, the free fields are just of the form $k(x_1, \ldots, x_n)$, where $x_1, \ldots, x_n$ are indeterminates. In the noncommutative case, you get something much more complicated, but free division rings can be realized as subrings of the noncommutative analogue of Laurent series.